The value of $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$ is

$\int_0^{\pi / 4} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$

If $f(x) = |x - 1|$,then $\int_0^2 {f(x)dx}$ is

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^2}}}{{\sec }^2}\frac{1}{{{n^2}}} + \frac{2}{{{n^2}}}{{\sec }^2}\frac{4}{{{n^2}}} + ..... + \frac{1}{n}{{\sec }^2}1} \right]$ equals

Let $f(x) = \begin{cases} \frac{1}{3}, & x \le \pi/2 \\ \frac{b(1-\sin x)}{(\pi-2x)^2}, & x > \pi/2 \end{cases}$. If $f$ is continuous at $x = \pi/2$, then the value of $\int_0^{3b-6} |x^2+2x-3| dx$ is:

The value of the integral $\int_{-\pi/4}^{\pi/4} \sin^4 x \, dx$ is